Please help, I'm finding difficulty in this math problem

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Zajie
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13 Sep 2015, 1:20 pm

I'm finding difficulty in Solving this :

H (x) = 1/ 4√x^2 - 5x



Rudin
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13 Sep 2015, 6:37 pm

I am extremely uncertain what you are asking. You have basically just stating a function, I'm sure what you want me to do.


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Kiriae
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14 Sep 2015, 8:20 am

Isn't it actually H(x)=- 4 3/4x (aka -4,75x)?
√x^2 = x because square power(^2) and square root(√) annihilate each other.

H (x) = 1/ 4√x^2 - 5x
H (x) = 1/ 4x - 5x
H (x) = - 4 3/4x
H (x)= -4,75x

In that case this if a line function.
y=0
0=-4,75x |:-4,75
x=0/-4,75=0

^ counting the x where function has value of 0 (y=0) aka where it crosses x line.
(point 0,0)

x=0
y=-4,75*0=0

^ counting the y value for x=0 aka where it crosess y line.
(point 0,0)

It crosses line the x and y lines in just one point: (0,0)

x=1
y=-4,75*1=-4,75
It passes point (1,-4,75)

As you put the 2 points (0,0), (1,-4,75) on the chart and connect them you will see it is a descending line function (I might call it wrong because I am translating the name from Polish).
It will be similar to this one but slightly more steep:
Image


Unless the whole thing is under 1, but then it should be written as H (x) = 1/(4√x^2 - 5x).
Then the answer would be:
H (x) = 1/(4√x^2 - 5x)
H (x) = 1/(4x - 5x)
H (x) = - 1/x

In that case:
y=0
0=-1/x |*x
0x=-1
0=-1 WRONG!! 0 isn't -1
It never crosses line x (no point with y=0)

x=0
y=-1/0 WRONG!! Never divide by 0!
It never crosses line y (no point with x=0)

x=1
y=-1/1
y=-1
point (1,-1)
x=-1
y=-1/-1
y=1
point (-1,1)
x=-2
y=-1/-2
y=0,5
point (-2, 0,5)
x=-0,5
y=-1/-0,5
y=2
point (-0,5,2)
and so on.

After putting enough points on chart (like this: http://www.matmana6.pl/zdjecia/szkola_s ... a_x_10.PNG) you will see the function is inverted hiperbole.
Similar to this one but with different values:
Image


I hope it helped. I might have made some mistakes because last time I had to deal with examples like this was 8 years ago. And I used Polish then.



Rudin
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14 Sep 2015, 5:00 pm

Kiriae wrote:
Isn't it actually H(x)=- 4 3/4x (aka -4,75x)?
√x^2 = x because square power(^2) and square root(√) annihilate each other.

H (x) = 1/ 4√x^2 - 5x
H (x) = 1/ 4x - 5x
H (x) = - 4 3/4x
H (x)= -4,75x

In that case this if a line function.
y=0
0=-4,75x |:-4,75
x=0/-4,75=0

^ counting the x where function has value of 0 (y=0) aka where it crosses x line.
(point 0,0)

x=0
y=-4,75*0=0

^ counting the y value for x=0 aka where it crosess y line.
(point 0,0)

It crosses line the x and y lines in just one point: (0,0)

x=1
y=-4,75*1=-4,75
It passes point (1,-4,75)

As you put the 2 points (0,0), (1,-4,75) on the chart and connect them you will see it is a descending line function (I might call it wrong because I am translating the name from Polish).
It will be similar to this one but slightly more steep:
Image


Unless the whole thing is under 1, but then it should be written as H (x) = 1/(4√x^2 - 5x).
Then the answer would be:
H (x) = 1/(4√x^2 - 5x)
H (x) = 1/(4x - 5x)
H (x) = - 1/x

In that case:
y=0
0=-1/x |*x
0x=-1
0=-1 WRONG!! 0 isn't -1
It never crosses line x (no point with y=0)

x=0
y=-1/0 WRONG!! Never divide by 0!
It never crosses line y (no point with x=0)

x=1
y=-1/1
y=-1
point (1,-1)
x=-1
y=-1/-1
y=1
point (-1,1)
x=-2
y=-1/-2
y=0,5
point (-2, 0,5)
x=-0,5
y=-1/-0,5
y=2
point (-0,5,2)
and so on.

After putting enough points on chart (like this: http://www.matmana6.pl/zdjecia/szkola_s ... a_x_10.PNG) you will see the function is inverted hiperbole.
Similar to this one but with different values:
Image


I hope it helped. I might have made some mistakes because last time I had to deal with examples like this was 8 years ago. And I used Polish then.


No. √x^2 = |x|.


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"God may not play dice with the universe, but something strange is going on with prime numbers."

-Paul Erdos

"There are two types of cryptography in this world: cryptography that will stop your kid sister from looking at your files, and cryptography that will stop major governments from reading your files."

-Bruce Schneider


izzeme
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15 Sep 2015, 6:09 am

Rudin wrote:
Kiriae wrote:
-snip-
Similar to this one but with different values:
Image


No. √x^2 = |x|.


true, then just flip the right part of the graph over the horizontal axis, the logic barely changes



Kiriae
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15 Sep 2015, 8:09 am

Rudin wrote:
No. √x^2 = |x|.

Thanks for pointing it out.
Looks like 8 years messes up with memory after all. :lol:



Adamantium
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15 Sep 2015, 9:36 am

Entertaining thread, complete with a twist ending for anyone who wasn't paying close attention through the main exposition.

I wonder if it answered the highly obscure question of the OP?



b9
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15 Sep 2015, 11:44 am

Zajie wrote:
I'm finding difficulty in Solving this :

H (x) = 1/ 4√x^2 - 5x


if (x) is not a subscript of H (but a multiplier), then H=-4.75

H*x= (1/4) * (√x^2) -5x
H*x= (.25*x) - 5x
h*x=-4.75*x
h=-4.75

it seems simple to me, but i am not sure if that is what you are asking.



slave
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04 Oct 2015, 8:48 pm

Zajie wrote:
I'm finding difficulty in Solving this :

H (x) = 1/ 4√x^2 - 5x


Ummm....so...did you read the responses to your inquiry?